2012 UK Puzzle Championship (2427 Aug 2012)
Re: 2012 UK Puzzle Championship (2427 Aug 2012)
Nick  I did this puzzle during the test, and the big thing I noticed with this puzzle was the interaction between the vertices and edges of a clued square cell being used as part of a triangular mine. So for example, if you rule out one vertex from a cell then there is only one possible way a mine can possibly go in an adjacent cell. I'm not in a position to provide a walkthrough, but using this approach I found relatively painless bifurcation routes through the puzzle. Although I do recall quite a lot of pain making sure there 15 rather than 14 mines used in the grid!

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Re: 2012 UK Puzzle Championship (2427 Aug 2012)
#7 & #8  CNotes
Labelling:
TL TC TR
ML MC MR
BL BC BR
Each digit must represent Tens(T) or Units(U).
For each row/col we should determine how many of each type we are expecting.
Remember always that the Units for each row/col must sum to 10 or 20.
CNote #1:
Start with top row, since it had 3 large numbers (any 2 will sum>10)
Cannot have more than 1xT. Cannot have 3xU > must contain T/U/U.
If TL is T, TC/TR are U, so TL = 57 (units in R1 sum to 10)
ML must be U, so BL = 16 (units in Col 1 sum to 10), and ML = 27.
One of BC/BR is U, other is T, so values would be 15/69 or 58/26.
If 58/26, MC is 45, and col 2 cannot sum to 100.
If 15/69, MC is 48, MR is 25. completing top row with 37/6 completes the grid satisfactorily. Done.
For completeness:
If TC was the T, TC=79, and we cannot complete col 2.
If TR was the T, TR=68, BR is U, MR=26 > BR=06, BL/BC = 19/75 or 41/53.
19/75 would force MC to be 48, and col 2 cannot be completed.
41/53 would force ML to be 74, and col 1 cannot be completed.
CNote #2:
Note again that none of the rows or columns sum to a multiple of 10, so all have at least one T
col1, row1 and row 3 are all must have 1xT and 2xU.
We can check these rows/cols to find where the T is, by seeing what its U value should be in each direction.
Either of TL/BL being T results in that cell being 79. If there had been a parity difference, we could have discounted them. Will come back to these if necessary later.
If TR is T, TR=79 and we cannot complete col 3, so TR is U. Now we know col 3 is T/U/U also.
since row 3 and col 3 are both T/U/U, we can test BR. If BR is T, it would need to be 81 to satisfy col 3, and 80 to satisfy row 3. Hence it is U, and MR = 25
If TL is T, TL=79, MC=61, ML=14, BL=7. BC=35, TC=4, TR=17, BR=58, check, and we're done.
For completeness:
If ML is T, BC=35, MC is U (row 2), TC=49, MC=16, and row 2 cannot be completed.
If BL is T, BL=79, TC=46, and we can't complete col 2.
Unique answer for both puzzles. 2nd definitely trickier than the 1st.
Labelling:
TL TC TR
ML MC MR
BL BC BR
Each digit must represent Tens(T) or Units(U).
For each row/col we should determine how many of each type we are expecting.
Remember always that the Units for each row/col must sum to 10 or 20.
CNote #1:
Start with top row, since it had 3 large numbers (any 2 will sum>10)
Cannot have more than 1xT. Cannot have 3xU > must contain T/U/U.
If TL is T, TC/TR are U, so TL = 57 (units in R1 sum to 10)
ML must be U, so BL = 16 (units in Col 1 sum to 10), and ML = 27.
One of BC/BR is U, other is T, so values would be 15/69 or 58/26.
If 58/26, MC is 45, and col 2 cannot sum to 100.
If 15/69, MC is 48, MR is 25. completing top row with 37/6 completes the grid satisfactorily. Done.
For completeness:
If TC was the T, TC=79, and we cannot complete col 2.
If TR was the T, TR=68, BR is U, MR=26 > BR=06, BL/BC = 19/75 or 41/53.
19/75 would force MC to be 48, and col 2 cannot be completed.
41/53 would force ML to be 74, and col 1 cannot be completed.
CNote #2:
Note again that none of the rows or columns sum to a multiple of 10, so all have at least one T
col1, row1 and row 3 are all must have 1xT and 2xU.
We can check these rows/cols to find where the T is, by seeing what its U value should be in each direction.
Either of TL/BL being T results in that cell being 79. If there had been a parity difference, we could have discounted them. Will come back to these if necessary later.
If TR is T, TR=79 and we cannot complete col 3, so TR is U. Now we know col 3 is T/U/U also.
since row 3 and col 3 are both T/U/U, we can test BR. If BR is T, it would need to be 81 to satisfy col 3, and 80 to satisfy row 3. Hence it is U, and MR = 25
If TL is T, TL=79, MC=61, ML=14, BL=7. BC=35, TC=4, TR=17, BR=58, check, and we're done.
For completeness:
If ML is T, BC=35, MC is U (row 2), TC=49, MC=16, and row 2 cannot be completed.
If BL is T, BL=79, TC=46, and we can't complete col 2.
Unique answer for both puzzles. 2nd definitely trickier than the 1st.
Re: 2012 UK Puzzle Championship (2427 Aug 2012)
Interesting, I used the same rules but in the opposite order to get a very different solution path.PuzzleScot wrote:#11 YinYang Different Neighbours.
I started by looking at the border. The two rectangles in the middle of the left side must be different colours. And the two squares adjacent to the topright corner must be different colours. So I coloured around the border (the top and tophalf of the left were pink, the right, bottom, and bottomofleft were green highlighter) leaving only the very topright cell empty. Then fill in the interior of the grid. At the end you'll find which colour is even and which is odd. Then enter the 1s,2s,3s,4s to get the answer key.

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Re: 2012 UK Puzzle Championship (2427 Aug 2012)
#14 Magic Minesweepers
Grid coordinates: columns ah, rows 18, top left is a1
1) From clue h1: g1/h2 are empty.
2) From the givens in e2/h7: e7 cannot contain either type of mine, and is empty.
3) From clue d1: at least one of c1/d2 is a mine.
4) From clues c4/c6: at least one of b3/c3/d3 is a mine.
5) Combining clue c2 with steps 3 and 4: EXACTLY one of c1/d2 is a mine, exactly one of b3/c3/d3 is a mine, b1/c2 are empty.
6) Combining clue d1 and step 5: e1 is a mine.
7) From given e2: e1 is type A. e3/e4/e5/e6/ are empty.
8) Combining step 5) with clues c4/d6: exactly one of b5/c5/d5 is a mine, b4/c4 are both mines, b6/d6/c7/d7 are empty.
(I think the solve became trivial a couple of steps ago, but for completeness...)
9) As b4/d4 are both mines: a4/f4/g4 are empty.
10) From clue f3: f2/g3 are mines. f3 is type A. a2/d2 are empty.
11) From clue d1: c1 is a mine of type B. a1/f1 are empty.
12) From clue g2: h3 is empty.
13) From clue e8: d8/f8 are mines. b8/c8/g8/h8 are empty. f6 is empty. d3/d5 are empty.
14) From clue f7: g6/g7 are empty.
15) From clue b7: a6/a7 are mines. a3 is empty.
16) From clue a5: b5 is empty.
17) From clue c6: c5 is a mine. c3 is empty.
18) From clue c4: b3 is a mine.
19) From clue f5: g5 is a mine. h5 is empty.
20) To get two mines in row 6: h6 is a mine.
21) All mine types are now forced by following the chains from the givens.
Grid coordinates: columns ah, rows 18, top left is a1
1) From clue h1: g1/h2 are empty.
2) From the givens in e2/h7: e7 cannot contain either type of mine, and is empty.
3) From clue d1: at least one of c1/d2 is a mine.
4) From clues c4/c6: at least one of b3/c3/d3 is a mine.
5) Combining clue c2 with steps 3 and 4: EXACTLY one of c1/d2 is a mine, exactly one of b3/c3/d3 is a mine, b1/c2 are empty.
6) Combining clue d1 and step 5: e1 is a mine.
7) From given e2: e1 is type A. e3/e4/e5/e6/ are empty.
8) Combining step 5) with clues c4/d6: exactly one of b5/c5/d5 is a mine, b4/c4 are both mines, b6/d6/c7/d7 are empty.
(I think the solve became trivial a couple of steps ago, but for completeness...)
9) As b4/d4 are both mines: a4/f4/g4 are empty.
10) From clue f3: f2/g3 are mines. f3 is type A. a2/d2 are empty.
11) From clue d1: c1 is a mine of type B. a1/f1 are empty.
12) From clue g2: h3 is empty.
13) From clue e8: d8/f8 are mines. b8/c8/g8/h8 are empty. f6 is empty. d3/d5 are empty.
14) From clue f7: g6/g7 are empty.
15) From clue b7: a6/a7 are mines. a3 is empty.
16) From clue a5: b5 is empty.
17) From clue c6: c5 is a mine. c3 is empty.
18) From clue c4: b3 is a mine.
19) From clue f5: g5 is a mine. h5 is empty.
20) To get two mines in row 6: h6 is a mine.
21) All mine types are now forced by following the chains from the givens.

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Re: 2012 UK Puzzle Championship (2427 Aug 2012)
#24 Triangle Minesweepers
Grid coordinates: columns AG, rows 17, top left is A1
Cell halves are TL, TR, BL, BR.
Any edgeadjacent group of three cells includes a total of 8 cell vertices, so cannot contain three nontouching triangles.
Hence, from clue B1: A2 and C1 contain mines (I initially mark this with a small central circle).
A mine edgeadjacent to a clue cell must touch that clue cell.
Hence, comparing clues C4/D5: E5/D6 contain mines. C3/B4 are empty.
Following this down to clue E6: F6/E7 are empty.
From clue G7: F7 contains a mine. It cannot touch E6, so is placed BR.
Clue G6 can not have any more touches: G5 is empty. Mines in E5/F5 would have to touch G6: F5 is empty.
If D1 contained a mine, C2 could not (no trios), and B2 would have to (clue B1). But then mines would touch in B2/C1. So D1 is empty.
We cannot now place 2 mines touching E1 without one of them touching F1: G1 is therefore empty.
From clues G2/G4: G3 must contain a mine, otherwise F2/F3/F4 would all contain mines.
From clues G2/G4: if F3 contained a mine, F2/F4 would be empty. But then, either G2 or G4 would not be touched by both F3 and G3. F3 is therefore empty.
From clues G2/G4 (one more time!): F2/F4 contain mines.
From clue B1: there is a mine in B2 or C2, and it must touch the corner of C2/B3, so a mine in B3 would have to be BL. But this would touch clue C4, and we already know that C4 must be touched a mine in D4 or C5. So B3 is empty.
I'm sure I can do better than this, but SOMEWHAT DODGY UNIQUENESS ASSUMPTION ALERT!
Given mines in F2/G3/F4, we cannot put a mine in E3. Why not? That set of cells has 12 vertices, so every vertex would be touched by a mine  but there are two mirrored ways of doing that. (F2:TL, E3:BL, F4:BR, G3:TR or F2:TR, E3:TL, F4:BL, G3:BR) So E3 is empty.
From clue E4: D4 contains a mine.
From clue C4: C5 is empty.
From clue A7: A6/B7 contain exactly one mine.
From clue B6: B5/C7 contain mines.
From mines D6/C7: D7 must be empty.
As clue C4 must be touched by mine D4, mine B5 must be BL.
From clue E1: D2/E2 contain exactly one mine. If D2, it touches clue D3. If E2, the E2/F2 pair touches D3. So a mine in C2 would have to be TL, but the mine in C1 forbids that. So C2 is empty.
From clue B1: B2 contains a mine. A3 is empty.
Mine A2 touches clue A1, so mine B2 does not. B2 is BR, A2 is TL, C1 is TR.
There are now five candidate cells for three remaining mines. D2/E2 contain no more than one, A6/B7 contain no more than one. A4 is a mine, TL.
Clue A5 is touched by mines A4/B5, so A6 is empty, B7 is a mine.
Mines B7/C7 will touch cell D6: mine D6 and, in turn, mines E5/F4/G3 are all TR. Then, mines F2/D4 are TL.
From mine F2, E2 must be empty. D2 must be the 15th mine, and must be BR.
Mines B5/D6/C7 will all touch clue C6, so mine B7 does not and is BL. Therefore mine C7 is TR.
A quick countup gives us a solution key of 4,6,3,2  as a crosscheck, this sums to 15, so we're done.
Grid coordinates: columns AG, rows 17, top left is A1
Cell halves are TL, TR, BL, BR.
Any edgeadjacent group of three cells includes a total of 8 cell vertices, so cannot contain three nontouching triangles.
Hence, from clue B1: A2 and C1 contain mines (I initially mark this with a small central circle).
A mine edgeadjacent to a clue cell must touch that clue cell.
Hence, comparing clues C4/D5: E5/D6 contain mines. C3/B4 are empty.
Following this down to clue E6: F6/E7 are empty.
From clue G7: F7 contains a mine. It cannot touch E6, so is placed BR.
Clue G6 can not have any more touches: G5 is empty. Mines in E5/F5 would have to touch G6: F5 is empty.
If D1 contained a mine, C2 could not (no trios), and B2 would have to (clue B1). But then mines would touch in B2/C1. So D1 is empty.
We cannot now place 2 mines touching E1 without one of them touching F1: G1 is therefore empty.
From clues G2/G4: G3 must contain a mine, otherwise F2/F3/F4 would all contain mines.
From clues G2/G4: if F3 contained a mine, F2/F4 would be empty. But then, either G2 or G4 would not be touched by both F3 and G3. F3 is therefore empty.
From clues G2/G4 (one more time!): F2/F4 contain mines.
From clue B1: there is a mine in B2 or C2, and it must touch the corner of C2/B3, so a mine in B3 would have to be BL. But this would touch clue C4, and we already know that C4 must be touched a mine in D4 or C5. So B3 is empty.
I'm sure I can do better than this, but SOMEWHAT DODGY UNIQUENESS ASSUMPTION ALERT!
Given mines in F2/G3/F4, we cannot put a mine in E3. Why not? That set of cells has 12 vertices, so every vertex would be touched by a mine  but there are two mirrored ways of doing that. (F2:TL, E3:BL, F4:BR, G3:TR or F2:TR, E3:TL, F4:BL, G3:BR) So E3 is empty.
From clue E4: D4 contains a mine.
From clue C4: C5 is empty.
From clue A7: A6/B7 contain exactly one mine.
From clue B6: B5/C7 contain mines.
From mines D6/C7: D7 must be empty.
As clue C4 must be touched by mine D4, mine B5 must be BL.
From clue E1: D2/E2 contain exactly one mine. If D2, it touches clue D3. If E2, the E2/F2 pair touches D3. So a mine in C2 would have to be TL, but the mine in C1 forbids that. So C2 is empty.
From clue B1: B2 contains a mine. A3 is empty.
Mine A2 touches clue A1, so mine B2 does not. B2 is BR, A2 is TL, C1 is TR.
There are now five candidate cells for three remaining mines. D2/E2 contain no more than one, A6/B7 contain no more than one. A4 is a mine, TL.
Clue A5 is touched by mines A4/B5, so A6 is empty, B7 is a mine.
Mines B7/C7 will touch cell D6: mine D6 and, in turn, mines E5/F4/G3 are all TR. Then, mines F2/D4 are TL.
From mine F2, E2 must be empty. D2 must be the 15th mine, and must be BR.
Mines B5/D6/C7 will all touch clue C6, so mine B7 does not and is BL. Therefore mine C7 is TR.
A quick countup gives us a solution key of 4,6,3,2  as a crosscheck, this sums to 15, so we're done.

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Re: 2012 UK Puzzle Championship (2427 Aug 2012)
#23 Crypted Fence (Decryption)
Columns AO, rows 16, top left is A1.
All clue letters appear multiple times, so there's no sneaky 4 in there  the clues represent 0, 1, 2, 3.
There are three diagonally adjacent C's in E5/F4/G3; C cannot represent 3.
If I4/I5/I6 were all 3s, J5 could not also be 3; U cannot represent 3.
A 3clue cannot be adjacent to a 0clue at the edge of a grid; P is adjacent to K(H1), C(I1) and U(C6), so P cannot represent 3.
Therefore K represents 3.
On the edge of the grid, K is adjacent to U(A1) and P(C1), so U and P do not represent 0.
Therefore C represents 0.
With 0 and 3 clues placed, we can mark in part of the loop around E6, noting that there is a line between F5/F6 and blank below F6.
If U were 1 and P was 2, the line around J5 would be bottom or left, and the line around I6 would be top or right; I6 bottom and left would be blank, so H6 bottom would also be blank.
That in turn would make it impossible to fulfil a 2clue in G6.
Therefore P represents 1, U represents 2.
The puzzle now collapses down to a standard (though still rather hard!) Fences puzzle. I suspect that the block of 2clues in the I3 area might have a forced solution making the remainder relatively straightforward, but I can't quite put my finger on it. Even though I've solved it twice now, I'm struggling to lay down a clear logical path, so I'll let that be.
Columns AO, rows 16, top left is A1.
All clue letters appear multiple times, so there's no sneaky 4 in there  the clues represent 0, 1, 2, 3.
There are three diagonally adjacent C's in E5/F4/G3; C cannot represent 3.
If I4/I5/I6 were all 3s, J5 could not also be 3; U cannot represent 3.
A 3clue cannot be adjacent to a 0clue at the edge of a grid; P is adjacent to K(H1), C(I1) and U(C6), so P cannot represent 3.
Therefore K represents 3.
On the edge of the grid, K is adjacent to U(A1) and P(C1), so U and P do not represent 0.
Therefore C represents 0.
With 0 and 3 clues placed, we can mark in part of the loop around E6, noting that there is a line between F5/F6 and blank below F6.
If U were 1 and P was 2, the line around J5 would be bottom or left, and the line around I6 would be top or right; I6 bottom and left would be blank, so H6 bottom would also be blank.
That in turn would make it impossible to fulfil a 2clue in G6.
Therefore P represents 1, U represents 2.
The puzzle now collapses down to a standard (though still rather hard!) Fences puzzle. I suspect that the block of 2clues in the I3 area might have a forced solution making the remainder relatively straightforward, but I can't quite put my finger on it. Even though I've solved it twice now, I'm struggling to lay down a clear logical path, so I'll let that be.
Re: 2012 UK Puzzle Championship (2427 Aug 2012)
#24 Triangle Minesweeper
In the contest I made an educated 'guess' and used some uniqueness arguments (if a triangle can rotate and satisfy the same clues then something's wrong).
However there is a nice logical path from start to finish that doesn't need those. I created this image yesterday but nickdeller has now written up a full walkthrough (he has been busy!), so I'll still show you my two halfway points as they might be helpful, but will only add a quick explanation as a lot of the logic is the same as his.
In this case blue marks indicate a mine, red marks mean no mine.
The top 3 only has three available corners, so each corner gets a different mine. In particular C1 can be drawn as a TRtriangle.
Now the top 2 only has two corners, so again mark in some blue lines and cross out G1.
The 1 in C4 and the 3 in D5 combine to give mines in D6, E5, and red marks in B3, B4, B5, C3.
Now the 2 in E6 is complete, so surround the rest with red.
This leaves a BRmine in F7. And cross out F5 and G5.
The 2 in G4 needs two mines in F3, F4, G3, but F2 is a mine and three mines cannot be adjacent, so rule out F3.
For the 3 in B6, it cannot touch two mines in the two cells above (B5, C5), and cannot touch two mines in the cells to the left and below (A6, B7), so there must be a mine in C7.
Now we have the lefthand grid image.
Look at E3. If there is a mine there then it pushes the F2 and F4 mines to their far corners, which leaves nowhere for G3. So E3 is empty. (this is where nickdeller used a clever uniqueness rule, but in our case I've ruled out the topleft corner of E3 already, and we can't fit 4 triangles into 11 corners)
D4 is a mine. C5 is not. B5 is a mine.
C2 must be empty because the 2 is filled, so A2 and B2 are filled.
D7 is empty, and we have the righthand grid image.
We now have 13 mines located. There are solutions for either 14 (=A6) or 15 (=A4 and B7) mines total. We need 15, and when you enter B7 you will trigger a chain of minelockingins, through C7, D6, E5, D4, F4, G3, F2, D2.
In the contest I made an educated 'guess' and used some uniqueness arguments (if a triangle can rotate and satisfy the same clues then something's wrong).
However there is a nice logical path from start to finish that doesn't need those. I created this image yesterday but nickdeller has now written up a full walkthrough (he has been busy!), so I'll still show you my two halfway points as they might be helpful, but will only add a quick explanation as a lot of the logic is the same as his.
In this case blue marks indicate a mine, red marks mean no mine.
The top 3 only has three available corners, so each corner gets a different mine. In particular C1 can be drawn as a TRtriangle.
Now the top 2 only has two corners, so again mark in some blue lines and cross out G1.
The 1 in C4 and the 3 in D5 combine to give mines in D6, E5, and red marks in B3, B4, B5, C3.
Now the 2 in E6 is complete, so surround the rest with red.
This leaves a BRmine in F7. And cross out F5 and G5.
The 2 in G4 needs two mines in F3, F4, G3, but F2 is a mine and three mines cannot be adjacent, so rule out F3.
For the 3 in B6, it cannot touch two mines in the two cells above (B5, C5), and cannot touch two mines in the cells to the left and below (A6, B7), so there must be a mine in C7.
Now we have the lefthand grid image.
Look at E3. If there is a mine there then it pushes the F2 and F4 mines to their far corners, which leaves nowhere for G3. So E3 is empty. (this is where nickdeller used a clever uniqueness rule, but in our case I've ruled out the topleft corner of E3 already, and we can't fit 4 triangles into 11 corners)
D4 is a mine. C5 is not. B5 is a mine.
C2 must be empty because the 2 is filled, so A2 and B2 are filled.
D7 is empty, and we have the righthand grid image.
We now have 13 mines located. There are solutions for either 14 (=A6) or 15 (=A4 and B7) mines total. We need 15, and when you enter B7 you will trigger a chain of minelockingins, through C7, D6, E5, D4, F4, G3, F2, D2.

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Re: 2012 UK Puzzle Championship (2427 Aug 2012)
Excellent walkthrough James. The images help of course, but I'm loving your annotation style and use of colour!

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Re: 2012 UK Puzzle Championship (2427 Aug 2012)
Time for one of the easy ones...
Puzzle #3  Tapa #1
Columns labelled leftright AH, rows labelled topbottom 18.
The cells around the 5 at B8 are all shaded: A7, B7, C7, A8, C8 are shaded.
C7 is shaded, so to complete the 1/1/1/1 at D6: C5, E5, E7 are shaded. D5, C6, E6, D7 are blank.
The 1/3 at G1 can only be achieved by: F1, H1, G2 and H2 are shaded, F2 is blank.
Since F2 is blank the 3/3 at E3 is achieved by: D2, E2, D3, F3, E4, F4 are shaded. D4 is blank. Also, F5 is blank.
The right and left hand sides can only connect via the bottom of the grid: D8 and E8 are shaded.
Similarly, the connection must pass to the right of the 4 at G6: H5, H6 and H7 are shaded.
F6 and F7 are therefore blank, and F8 and G8 are shaded.
If G3 was shaded, H3 and G4 would be blank, but this would cut off the shaded section in that area. therefore:
G3 is blank. H3, G4, H4 are shaded and G5 is blank.
To complete the 4 at G6: G7 is shaded. H8 is blank.
To connect F1: E1 is shaded. D1 is blank.
D2 or D3 must connect via either C2 or C3 but only one of C2/C3 can be shaded. If C2 is shaded, with C3 blank, then the shading cannot reach down past the 4 at B3, therefore: C3 is shaded, C2 is blank. A4, B4, C4 are shaded. A2, B2, C2 are blank.
B5 is blank.
To complete the connection: A5 and A6 are shaded.
The remaining cells at A1, B1, C1 and B6 are blank.
Puzzle #3  Tapa #1
Columns labelled leftright AH, rows labelled topbottom 18.
The cells around the 5 at B8 are all shaded: A7, B7, C7, A8, C8 are shaded.
C7 is shaded, so to complete the 1/1/1/1 at D6: C5, E5, E7 are shaded. D5, C6, E6, D7 are blank.
The 1/3 at G1 can only be achieved by: F1, H1, G2 and H2 are shaded, F2 is blank.
Since F2 is blank the 3/3 at E3 is achieved by: D2, E2, D3, F3, E4, F4 are shaded. D4 is blank. Also, F5 is blank.
The right and left hand sides can only connect via the bottom of the grid: D8 and E8 are shaded.
Similarly, the connection must pass to the right of the 4 at G6: H5, H6 and H7 are shaded.
F6 and F7 are therefore blank, and F8 and G8 are shaded.
If G3 was shaded, H3 and G4 would be blank, but this would cut off the shaded section in that area. therefore:
G3 is blank. H3, G4, H4 are shaded and G5 is blank.
To complete the 4 at G6: G7 is shaded. H8 is blank.
To connect F1: E1 is shaded. D1 is blank.
D2 or D3 must connect via either C2 or C3 but only one of C2/C3 can be shaded. If C2 is shaded, with C3 blank, then the shading cannot reach down past the 4 at B3, therefore: C3 is shaded, C2 is blank. A4, B4, C4 are shaded. A2, B2, C2 are blank.
B5 is blank.
To complete the connection: A5 and A6 are shaded.
The remaining cells at A1, B1, C1 and B6 are blank.

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Re: 2012 UK Puzzle Championship (2427 Aug 2012)
#17  Wittgenstein Briquet.
Cols AJ, rows 110. TL = A1
Due to 1/3 combo at D3/E2, C3/D4 out & E1/F3 ae in.
Only one way to satisfy the 1 at D3  block B2D2. E3 is out.
Remembering that all empty cells (including cells with clues) are contiguously connect, A1 can only be connected if A2&A3 are out.
Now B3 cannot be a block, so A4/B5/C4 are blocks. A4 block extends to A6, C4 to C6, so B5 extends to B7.
If B8 was a block, we're left with 4 isolated cells in col A, so B8 is out, C7/C9/D8 are all blocks.
C7 extends to E7. C9 is horizontal, but can't extend to A9 (enclosed area), so must include D9 at least.
This forces D8 to extend to F8. B10E10 cannot be blocks, to allow BL area out to connect with the other unused cells.
F9 and D5 cannot be part of any block, so are out.
To complete the 1 at A1, we need a block from B1D1. E1 extends to G1.
That's the easy part done.
If F1 was part of a block, it would extend to F5; To complete the 2 at E5, we'd need a block from E4G4. Now we cannot satisfy the 3 at G3 without enclosing the TL unused cells, so F1 is out. Likewise, if G8 was part of a block, we cannot satisfy the BR 2s without enclosing cells, so G8 is out too.
So, for the 2 at G9,we have a block from F10H10; E9 is out (to connect the BL area) and B9 is in. The rest of col A (710) is out, otherwise, unconnected area is formed. H9 is in, but can't go horizontally, or would block off BR corner, so we have a block from H9H7. G7 is satisfied, so G6 is out.
If G4 is out, satisfying the 2 at E5 encloses the TL region, so G4 is in (and is horizontal).
If F2 extends down, again, the 2 at E5 encloses TL region, so F2 extends to H2.
Again, if F3 is in, it extends down, causing a conflict with that 2 and the TL region, so F3 is out.
H3 is in. If it extends to J3, we have 5 enclosed cells in TR corner, so extends to H5.
G4 extends to E4, and we have a block from D6F6, with F5/G5/H6/I6 all empty.
2 at J4 can only be satified with J1J3 and J5J7.
If I9 was in, the block from I7I9 would enclose BR area, so block is at J8J10, with I9 empty.
With H9 empty, all of col I is empty too, for all the empty cells to connect. Done.
Note that often with this puzzle type, there are necessarily blocks that do not touch any clues!
Cols AJ, rows 110. TL = A1
Due to 1/3 combo at D3/E2, C3/D4 out & E1/F3 ae in.
Only one way to satisfy the 1 at D3  block B2D2. E3 is out.
Remembering that all empty cells (including cells with clues) are contiguously connect, A1 can only be connected if A2&A3 are out.
Now B3 cannot be a block, so A4/B5/C4 are blocks. A4 block extends to A6, C4 to C6, so B5 extends to B7.
If B8 was a block, we're left with 4 isolated cells in col A, so B8 is out, C7/C9/D8 are all blocks.
C7 extends to E7. C9 is horizontal, but can't extend to A9 (enclosed area), so must include D9 at least.
This forces D8 to extend to F8. B10E10 cannot be blocks, to allow BL area out to connect with the other unused cells.
F9 and D5 cannot be part of any block, so are out.
To complete the 1 at A1, we need a block from B1D1. E1 extends to G1.
That's the easy part done.
If F1 was part of a block, it would extend to F5; To complete the 2 at E5, we'd need a block from E4G4. Now we cannot satisfy the 3 at G3 without enclosing the TL unused cells, so F1 is out. Likewise, if G8 was part of a block, we cannot satisfy the BR 2s without enclosing cells, so G8 is out too.
So, for the 2 at G9,we have a block from F10H10; E9 is out (to connect the BL area) and B9 is in. The rest of col A (710) is out, otherwise, unconnected area is formed. H9 is in, but can't go horizontally, or would block off BR corner, so we have a block from H9H7. G7 is satisfied, so G6 is out.
If G4 is out, satisfying the 2 at E5 encloses the TL region, so G4 is in (and is horizontal).
If F2 extends down, again, the 2 at E5 encloses TL region, so F2 extends to H2.
Again, if F3 is in, it extends down, causing a conflict with that 2 and the TL region, so F3 is out.
H3 is in. If it extends to J3, we have 5 enclosed cells in TR corner, so extends to H5.
G4 extends to E4, and we have a block from D6F6, with F5/G5/H6/I6 all empty.
2 at J4 can only be satified with J1J3 and J5J7.
If I9 was in, the block from I7I9 would enclose BR area, so block is at J8J10, with I9 empty.
With H9 empty, all of col I is empty too, for all the empty cells to connect. Done.
Note that often with this puzzle type, there are necessarily blocks that do not touch any clues!

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Re: 2012 UK Puzzle Championship (2427 Aug 2012)
I'd say the opposite, actually  if the rules don't define the total number of blocks, as here, then a block that touches no clues could be removed without anything breaking. Am I missing something?PuzzleScot wrote:#17  Wittgenstein Briquet.
Note that often with this puzzle type, there are necessarily blocks that do not touch any clues!

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Re: 2012 UK Puzzle Championship (2427 Aug 2012)
Yes. All unoccupied cells must be contiguously connected.nickdeller wrote:Am I missing something?
If there was a block of 3 cells in the BL of that grid that could not be contiguous with the rest of the unoccupied cells whilst satisfying the remaining clues near the bottom, then it is perfectly acceptable to fill that space with an extra 3x1 block, so that doesn't become necessary. I haven't checked, but I'm sure that situation arose on (at least) one of of the OPAC WB puzzles...

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 Joined: Tue 29 Jun, 2010 11:41 am
Re: 2012 UK Puzzle Championship (2427 Aug 2012)
Ah, got it. Yes, that clicks now  as a concrete example, if there are blocks in A2/B2/C2 and D1/D2/D3, then there must be a block in A1/B1/C1. I think I might allow myself a little at overlooking that case!PuzzleScot wrote:Yes. All unoccupied cells must be contiguously connected.nickdeller wrote:Am I missing something?
If there was a block of 3 cells in the BL of that grid that could not be contiguous with the rest of the unoccupied cells whilst satisfying the remaining clues near the bottom, then it is perfectly acceptable to fill that space with an extra 3x1 block, so that doesn't become necessary. I haven't checked, but I'm sure that situation arose on (at least) one of of the OPAC WB puzzles...
Re: 2012 UK Puzzle Championship (2427 Aug 2012)
#15  Skyscrapers
With thanks to Serkan  Part of my problem for a good while now is that I get stuck at the pivot point of a problem, and it was no different with this one. After the initial solve, there is one key step in this problem  and I could not crack it  during and after the competition.
Row 4  can "complete" to 6 in the fifth column  with 1/2, 2/3, 3/4, 4/5 in the first four columns
Column 6  can "complete" to 6 in the fifth row  with 1/2, 2/3, 3/4, 4/5 in the first four rows [6 cannot be in last cell due to 2/3 conflict in bottom LHC which would need 5 there and this is impossible]. Then complete with 1, 2 in first two rows due to 3 clue at right end of second row
Row 4  can now complete columns 13 with 1,2,3
Pivot point:
Row 2  you cannot put a 6 in column 2, as then you would need a 1 in column 1 (impossible  already in row 4), and you cannot put a 5 in column 2 as 2 at top of column 2 does not permit this  solution would be 1 or 3 (depending on where you put a 6). Therefore, row 2 must be 6,1,5,4,3,2 [row 4 now 1,2,3,5,6,4]
Column 4  you cannot put a 6 at the bottom  row 6, due to 3 clue at top, and given 4,5 in rows 2,4 [first row 2/3], therefore 6 in row 3
Column 2/3  the remaining 6s can be added, row 6,1 respectively  then 5 at top of column 2, and by inspection 4/3 in rows 3/5
Column 3  4 must go in row 5 due to 4 clue at bottom
Column 5  5 cannot go in row 5 due to 2 clue at bottom, therefore must go in row 3, leaving remaining 5 to go in column 1 row 5
Row 3  1 cannot go in column 1, therefore, by inspection in column 3
Column 5  1 cannot go in row 6, due to 3 clue at bottom, therefore in row 5, and therefore, remaining 1 in row 6 in column 4
Column 4  row 5 must be 2, making row 1 a 3. Trivial to complete from this point.
With thanks to Serkan  Part of my problem for a good while now is that I get stuck at the pivot point of a problem, and it was no different with this one. After the initial solve, there is one key step in this problem  and I could not crack it  during and after the competition.
Row 4  can "complete" to 6 in the fifth column  with 1/2, 2/3, 3/4, 4/5 in the first four columns
Column 6  can "complete" to 6 in the fifth row  with 1/2, 2/3, 3/4, 4/5 in the first four rows [6 cannot be in last cell due to 2/3 conflict in bottom LHC which would need 5 there and this is impossible]. Then complete with 1, 2 in first two rows due to 3 clue at right end of second row
Row 4  can now complete columns 13 with 1,2,3
Pivot point:
Row 2  you cannot put a 6 in column 2, as then you would need a 1 in column 1 (impossible  already in row 4), and you cannot put a 5 in column 2 as 2 at top of column 2 does not permit this  solution would be 1 or 3 (depending on where you put a 6). Therefore, row 2 must be 6,1,5,4,3,2 [row 4 now 1,2,3,5,6,4]
Column 4  you cannot put a 6 at the bottom  row 6, due to 3 clue at top, and given 4,5 in rows 2,4 [first row 2/3], therefore 6 in row 3
Column 2/3  the remaining 6s can be added, row 6,1 respectively  then 5 at top of column 2, and by inspection 4/3 in rows 3/5
Column 3  4 must go in row 5 due to 4 clue at bottom
Column 5  5 cannot go in row 5 due to 2 clue at bottom, therefore must go in row 3, leaving remaining 5 to go in column 1 row 5
Row 3  1 cannot go in column 1, therefore, by inspection in column 3
Column 5  1 cannot go in row 6, due to 3 clue at bottom, therefore in row 5, and therefore, remaining 1 in row 6 in column 4
Column 4  row 5 must be 2, making row 1 a 3. Trivial to complete from this point.
Re: 2012 UK Puzzle Championship (2427 Aug 2012)
Don't worry, I'll have one too , I was sure your objection was fine!nickdeller wrote:I think I might allow myself a little at overlooking that case!
Re: 2012 UK Puzzle Championship (2427 Aug 2012)
Puzzle #4: Tapa 2
1. Edge effects. There are quite a few things to fill in at the start. You can completely fill in the 22 at r6c1, the 3 at r1c10.
You can fill in the orthogonally adjacent squares to both edge 13's, at r1c3 and r10c8. You can fill in the 3 "inland" squares on
both edge 4's at r5c10 and r10c5. You can fill in r9c2 from the 2 in the corner, and r7c4 from the opposite symmetry rule for 33's.
2. Lucky Sevens. So, the two 7's near the top of the puzzle are an obvious point of interest. A little thought shows that
you must have a total of 2 empty cells in all the cells surrounding those two 7's, but one of them must be in r2c57 (because of
the 2) and the other must be in r4c68 (because of the 22). All other cells touching those 7's (including the 2 cells in r3 between
the 7's) can be shaded in. We can also mark the cells in r1c5 and r1c7 as empty.
3. Path Busting. After marking a few empty cells from the "no 2x2 blocks" rule, we see that we have a cell in r1c2 that must
find an escape to the rest of the puzzle. We send it out to the left and down c1 all the way to connect to r5c1. From there it
must head to the right filling r5c3 to complete the 22 at r4c3. It has to head south next, so it heads down c3 to r6c3. This allows
us to completely fill in both the 24 and the 33 nearby. And we still have only one way to send this part of the wall  it goes left
along r7 to r7c1, then down c1 to r9c1, then right all the way to r9c8. If the path continued up through r8c7, it would dead end.
So that cell is empty, and it goes up in r8c8, then across to r8c10. Some loose ends we've created: finish the 4 in r10c5, the
"4" from the 14 in r8c6, and send the wall fragment in r10c9 around the corner to connect to the other part of the wall.
4. Stuck in the Middle With Twos. A little though about the 22 at r5c7 shows that there is only one way to complete it that
is consistent with no 2x2 blocks and the 121 nearby. That's filling in r47,8 and r6c6,7. That allows us to fill in the rest of
the 7's as well, and complete the 14 in r8c6 by filling in r7c6. The only way to connect the top right of the puzzle to the rest
is to send it around the 4 from r6c10 to r8c10. That should finish the puzzle.
1. Edge effects. There are quite a few things to fill in at the start. You can completely fill in the 22 at r6c1, the 3 at r1c10.
You can fill in the orthogonally adjacent squares to both edge 13's, at r1c3 and r10c8. You can fill in the 3 "inland" squares on
both edge 4's at r5c10 and r10c5. You can fill in r9c2 from the 2 in the corner, and r7c4 from the opposite symmetry rule for 33's.
2. Lucky Sevens. So, the two 7's near the top of the puzzle are an obvious point of interest. A little thought shows that
you must have a total of 2 empty cells in all the cells surrounding those two 7's, but one of them must be in r2c57 (because of
the 2) and the other must be in r4c68 (because of the 22). All other cells touching those 7's (including the 2 cells in r3 between
the 7's) can be shaded in. We can also mark the cells in r1c5 and r1c7 as empty.
3. Path Busting. After marking a few empty cells from the "no 2x2 blocks" rule, we see that we have a cell in r1c2 that must
find an escape to the rest of the puzzle. We send it out to the left and down c1 all the way to connect to r5c1. From there it
must head to the right filling r5c3 to complete the 22 at r4c3. It has to head south next, so it heads down c3 to r6c3. This allows
us to completely fill in both the 24 and the 33 nearby. And we still have only one way to send this part of the wall  it goes left
along r7 to r7c1, then down c1 to r9c1, then right all the way to r9c8. If the path continued up through r8c7, it would dead end.
So that cell is empty, and it goes up in r8c8, then across to r8c10. Some loose ends we've created: finish the 4 in r10c5, the
"4" from the 14 in r8c6, and send the wall fragment in r10c9 around the corner to connect to the other part of the wall.
4. Stuck in the Middle With Twos. A little though about the 22 at r5c7 shows that there is only one way to complete it that
is consistent with no 2x2 blocks and the 121 nearby. That's filling in r47,8 and r6c6,7. That allows us to fill in the rest of
the 7's as well, and complete the 14 in r8c6 by filling in r7c6. The only way to connect the top right of the puzzle to the rest
is to send it around the 4 from r6c10 to r8c10. That should finish the puzzle.

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 Joined: Thu 01 Sep, 2011 5:04 pm
Re: 2012 UK Puzzle Championship (2427 Aug 2012)
OK. I'm finally getting round to looking at these (it's been a busy few weeks) having been away at the August Bank Holiday. Can you unpack "nothing can occupy F2" please? It can't be a numbered block, but I can't see any reason why it can't be a different, unnumbered block?PuzzleScot wrote:#18. Tren.
This one seems to have caused more difficulties than most, so I'll give this a bash.
4 at F5 is vertical. Note 3 at F1 is horizontal, and nothing can occupy F2. Therefore this has to be a size 3: F3/4/5.
Thanks  and sorry to be dense.
Alison

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 Joined: Tue 29 Jun, 2010 11:41 am
Re: 2012 UK Puzzle Championship (2427 Aug 2012)
The rule clarification given here seems not to have been added to the master list in the top post, but answers the question  in Tren, there can't be an unnumbered block.BohemianCoast wrote:OK. I'm finally getting round to looking at these (it's been a busy few weeks) having been away at the August Bank Holiday. Can you unpack "nothing can occupy F2" please? It can't be a numbered block, but I can't see any reason why it can't be a different, unnumbered block?PuzzleScot wrote:#18. Tren.
This one seems to have caused more difficulties than most, so I'll give this a bash.
4 at F5 is vertical. Note 3 at F1 is horizontal, and nothing can occupy F2. Therefore this has to be a size 3: F3/4/5.
 furudo.erika
 Posts: 76
 Joined: Wed 10 Aug, 2011 11:55 am
Re: 2012 UK Puzzle Championship (2427 Aug 2012)
I did not attempt it as I was unfamiliar with the type. Where can I find more YinYang puzzles? Google is failing me.PuzzleScot wrote:#11 YinYang Different Neighbours.
I'm surprised how few people attempted this.
It can be solved as a YinYang, using Odd/Even markers, then all the values filled in at the end.

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 Location: Edinburgh, Scotland
Re: 2012 UK Puzzle Championship (2427 Aug 2012)
The first one I ever remember seeing was on PQRST 13: http://www.otuzoyun.com/pqrst/pqrst13.htmlfurudo.erika wrote:Where can I find more YinYang puzzles?
Sometimes it is unhelpfully called "Black and White", which I don't imagine being of any use to google searches!
There are a couple attached that appeared in French qualifying contests in 2004 & 2009. I don't know if the original links still work.
 Attachments

 Yin_Yang 1.pdf
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 Yin Yang 2.pdf
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